The given differential equation can be written as:
(1+sinx)dy+(y+1)cosxdx=0
d((y+1)(1+sinx))=0
Integrating both sides:
(y+1)(1+sinx)=C
Given y(0)=0, substituting x=0 and y=0:
(0+1)(1+sin0)=C⇒C=1
The equation of the curve is (y+1)(1+sinx)=1
Since the curve passes through (α,2−1), substituting x=α and y=2−1:
(2−1+1)(1+sinα)=1
21(1+sinα)=1
1+sinα=2
sinα=1
α=2π
Answer: 2π